HW7sol - π 3 f changes from negative to positive so f has...

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M408C, Homework 7 — Solution Instructor: Guoliang Wu 1. Find the limit lim x →-∞ 9 x 6 - x x 3 + 1 . Solution: lim x →-∞ 9 x 6 - x x 3 + 1 = lim x →-∞ 9 x 6 - x/x 3 ( x 3 + 1) /x 3 = lim x →-∞ 9 x 6 - x/ ( - x 3 ) · ( - 1) 1 + 1 /x 3 = lim x →-∞ q 9 x 6 - x x 6 · ( - 1) 1 + 1 /x 3 , because - x 3 = x 6 > 0 = lim x →-∞ - p 9 - 1 /x 5 1 + 1 /x 3 = - 3 . 2. Use the guidelines on PP243–245 to sketch the curve y = sin x 2 + cos x Solution: (a) Domain: all x . (b) y -intercept: f (0) = 0; x -intercepts: f ( x ) = 0 x = 0 , ± π, ± 2 π, ± 3 π, · · · (c) It’s not even/odd function, but it is a periodic function with period 2 π . We only need to consider [0 , 2 π ]. (d) No asymptotes. (e) Use quotient rule: y 0 = 1 + 2 cos x (2 + cos x ) 2 . Thus, f 0 ( x ) > 0 when 1 + 2 cos x > 0 cos x > - 1 2 0 < x < 2 π 3 or 4 π 3 < x < 2 π. So f is decreasing on ( 2 π 3 , 4 π 3 ) and increasing on (0 , 2 π 3 ) ( 4 π 3 , 2 π ). (f) From the first derivative test, at 2 π 3 , f 0 changes from positive to negative, so f has a local maximum and f ( 2 π 3 ) = 1
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Unformatted text preview: π 3 , f changes from negative to positive, so f has a local minimum and f ( 4 π 3 ) =-1 √ 3 . 1 (g) Use quotient rule again to find f 00 ( x ) = 2 sin x (cos x-1) (2 + cos x ) 3 . Because the denomenator > 0, and cos x-1 < 0, we know f 00 ( x ) > if sin x < 0, that is, π < x < 2 π . So f is concave up on ( π, 2 π ) and concave down on (0 ,π ). And f has an inflection point at ( π, 0) and (0 , 0). (h) Use all the above information, we sketch the graph of the function restricted to [0 , 2 π ]. 1 2 3 4 5 6-0.6-0.4-0.2 0.2 0.4 0.6 We then extended it using periodicity, to complete the graph.-10-5 5 10-0.6-0.4-0.2 0.2 0.4 0.6 2...
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